The perimeter of right angled triangle is $12$ cm and its hypotenuse is of length $5$ cm. Find the other two sides and calculate its area. Verify the result using heron' formula.

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Solution

$a + b + 5 = 12$

$a + b = 7 \Rightarrow a = 7 - b$ ...(1)

$a^{2} + b^{2} = 5^{2}$

$a^{2} + b^{2} = 25$ ...(2)

From (1) and (2)

$(7 - b)^{2} + b^{2} = 25$

$49 + b^{2} - 14 b + b^{2} = 25$

$2 b^{2} - 14 b + 24 = 0$

$b^{2} - 7 b + 12 = 0$

$(b - 4) (b - 3) = 0$

$\Rightarrow b = 4, 3$

$a = 7 - b \Rightarrow a = 3, 4$

Area $= \frac{1}{2} \times 4 \times 3 = 6$ sq. units

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The perimeter of right angled triangle is 12 cm and its hypotenuse is of length 5 cm. Find the other two sides and calculate its area. Verify the result using heron' formula.

a+b+5=12a+b=7⇒a=7−b...(1)a2+b2=52a2+b2=25...(2)From (1) and (2)(7−b)2+b2=2549+b2−14b+b2=252b2−14b+24=0b2−7b+12=0(b−4)(b−3)=0⇒b=4,3a=7−b⇒a=3,4Area=12×4×3=6sq. units

The perimeter of right angled triangle is $12$ cm and its hypotenuse is of length $5$ cm. Find the other two sides and calculate its area. Verify the result using heron' formula.